Questions: There are 75,000 live bacteria present in a culture in a flask. When an antibiotic is added to the culture, the number of live bacteria is reduced as shown by the equation. Approximately how many hours have passed when there are 1500 bacteria left alive? 1500=75,000 e^-0.1733 t A. 22.1 hours B. 22.3 hours C. 22.6 hours D. 22.9 hours

Solution Steps

To solve for the time \( t \) when there are 1500 bacteria left, we need to solve the given exponential decay equation for \( t \). The equation is: \[ 1500 = 75,000 e^{-0.1733 t} \]

1. Divide both sides by 75,000 to isolate the exponential term.
2. Take the natural logarithm (ln) of both sides to solve for \( t \).
3. Rearrange the equation to solve for \( t \).

We start with the equation representing the decay of bacteria: \[ 1500 = 75000 e^{-0.1733 t} \]

Dividing both sides by \( 75000 \) gives: \[ \frac{1500}{75000} = e^{-0.1733 t} \] This simplifies to: \[ 0.02 = e^{-0.1733 t} \]

Taking the natural logarithm of both sides, we have: \[ \ln(0.02) = -0.1733 t \] Calculating \( \ln(0.02) \) yields approximately \( -3.9120 \): \[ -3.9120 = -0.1733 t \]

Rearranging the equation to solve for \( t \): \[ t = \frac{-3.9120}{-0.1733} \approx 22.5737 \] Rounding to four significant digits, we find: \[ t \approx 22.57 \]

Final Answer